The scalar product of the vector hat{i}+hat{j | vedclass

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(A) Let $\vec{a} = \hat{i}+\hat{j}+\hat{k}$.Let $\vec{b} = 2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\vec{c} = \lambda \hat{i}+2 \hat{j}+3 \hat{k}$.The sum

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(A) Let $\vec{a} = \hat{i}+\hat{j}+\hat{k}$.Let $\vec{b} = 2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\vec{c} = \lambda \hat{i}+2 \hat{j}+3 \hat{k}$.The sum of the vectors is $\vec{v} = \vec{b} + \vec{c} = (2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}$.The unit vector along $\vec{v}$ is $\hat{v} = \frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{(2+\lambda)^2+6^2+(-2)^2}} = \frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{\lambda^2+4 \lambda+44}}$.The scalar product of $\vec{a}$ and $\hat{v}$ is $1$:$\vec{a} \cdot \hat{v} = 1$$\frac{(1)(2+\lambda) + (1)(6) + (1)(-2)}{\sqrt{\lambda^2+4 \lambda+44}} = 1$$\frac{\lambda+6}{\sqrt{\lambda^2+4 \lambda+44}} = 1$$\lambda+6 = \sqrt{\lambda^2+4 \lambda+44}$Squaring both sides:$(\lambda+6)^2 = \lambda^2+4 \lambda+44$$\lambda^2+12 \lambda+36 = \lambda^2+4 \lambda+44$$8 \lambda = 8$$\lambda = 1$.

The scalar product of the vector $\hat{i}+\hat{j}+\hat{k}$ with a unit vector along the sum of the vectors $2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\lambda \hat{i}+2 \hat{j}+3 \hat{k}$ is equal to $1$. Then the value of $\lambda$ is:

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